Complex Number Question =[ (1 Viewer)

[lilainjel]

Prove that 1+z+z^2+z^3 = (1-z^4)/(1-z)
Use this result and z=cos @ + isin @ to prove that:

1 + cos@ + cos2@ + cos3@ = (1/2)[1+(sin(7@/2))/(sin@/2)]

I can prove the first bit, but completely off the deep end with the second part.

 

[Arowana21]

sub in your z value and then get the real numbers

 

[lilainjel]

I couldn't break it down from there. I ended up with a whole assortment of cubes and double/triple angles and etc.

 

[Trebla]

1 + z + z² + z³ is a geometric series with a = 1 and r = z
Hence 1 + z + z² + z³ = (1 - z⁴)/(1 - z)
Let z = cis @
1 + cis @ + cis 2@ + cis 3@ = (1 - cis 4@)/(1 - cis @) by DeMoivre's theorem
= (1 - cos 4@ - isin 4@)(1 - cos @ + isin @) / (1 - cos @ + isin @)(1 - cos @ - isin @)
= [(1 - cos 4@)(1 - cos @) + isin @(1 - cos4@) - isin4@(1 - cos @) - i²sin @sin 4@] / [(1 - cos @)² + sin²@]
= [(1 - cos 4@) (1 - cos @) + sin @sin 4@ + i{sin @(1 - cos4@) - sin4@(1 - cos @)}] / [1 - 2cos @ + cos²@ + sin²@]
= [(1 - cos 4@) (1 - cos @) + sin @sin 4@ + i{sin @(1 - cos4@) - sin4@(1 - cos @)}] / [2 - 2cos @]
Equating real parts:
1 + cos @ + cos 2@ + cos 3@ = [(1 - cos 4@) (1 - cos @) + sin @sin 4@]/ [2 - 2cos @]
= [1 - cos @ - cos 4@ + cos @cos4@ + sin @sin4@] / [2 - 2cos @]
= [1 - cos @ - cos 4@ + cos (4@ - @)] / [2 + 2cos @]
= [1 - cos @ + cos 3@ - cos 4@] / 2[1 - cos @]

{but cos 3@ - cos 4@ = cos (7@/2 - @/2) - cos (7@/2 + @/2)
= cos 7@/2 cos @/2 + sin 7@/2 sin @/2 - cos 7@/2 cos @/2 + sin 7@/2 sin @/2
= 2sin 7@/2 sin @/2}

= [1 - cos @ + 2sin 7@/2 sin @/2] / 2[1 - cos @]
= [1 - (1 - 2sin² @/2) + 2sin 7@/2 sin @/2] / 2[1 - (1 - 2sin² @/2)]
= [2sin² @/2 + 2sin 7@/2 sin @/2] / 4sin² @/2
= [sin @/2 + sin 7@/2] / 2sin @/2
= 1/2 + sin 7@/2 / 2sin @/2
= 1/2[1 + (sin 7@/2 / sin @/2)]

 

[shaon0]

1 + z + z² + z³ is a geometric series with a = 1 and r = z
Hence 1 + z + z² + z³ = (1 - z⁴)/(1 - z)
Let z = cis @
1 + cis @ + cis 2@ + cis 3@ = (1 - cis 4@)/(1 - cis @) by DeMoivre's theorem
= (1 - cos 4@ - isin 4@)(1 - cos @ + isin @) / (1 - cos @ + isin @)(1 - cos @ - isin @)
= [(1 - cos 4@)(1 - cos @) + isin @(1 - cos4@) - isin4@(1 - cos @) - i²sin @sin 4@] / [(1 - cos @)² + sin²@]
= [(1 - cos 4@) (1 - cos @) + sin @sin 4@ + i{sin @(1 - cos4@) - sin4@(1 - cos @)}] / [1 - 2cos @ + cos²@ + sin²@]
= [(1 - cos 4@) (1 - cos @) + sin @sin 4@ + i{sin @(1 - cos4@) - sin4@(1 - cos @)}] / [2 - 2cos @]
Equating real parts:
1 + cos @ + cos 2@ + cos 3@ = [(1 - cos 4@) (1 - cos @) + sin @sin 4@ + i{sin @(1 - cos4@) - sin4@(1 - cos @)}] / [2 - 2cos @]
= [1 - cos @ - cos 4@ + cos @cos4@ + sin @sin4@] / [2 - 2cos @]
= [1 - cos @ - cos 4@ + cos (4@ - @)] / [2 + 2cos @]
= [1 - cos @ + cos 3@ - cos 4@] / 2[1 - cos @]

{but cos 3@ - cos 4@ = cos (7@/2 - @/2) - cos (7@/2 + @/2)
= cos 7@/2 cos @/2 + sin 7@/2 sin @/2 - cos 7@/2 cos @/2 + sin 7@/2 sin @/2
= 2sin 7@/2 sin @/2}

= [1 - cos @ + 2sin 7@/2 sin @/2] / 2[1 - cos @]
= [1 - (1 - 2sin² @/2) + 2sin 7@/2 sin @/2] / 2[1 - (1 - 2sin² @/2)]
= [2sin² @/2 + 2sin 7@/2 sin @/2] / 4sin² @/2
= [sin @/2 + sin 7@/2] / 2sin @/2
= 1/2 + sin 7@/2 / 2sin @/2
= 1/2[1 + (sin 7@/2 / sin @/2)]

Are you a genius? Seriously. I got upto equating reals.

 

[lilainjel]

Thank You!